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Question
Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx
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Solution
Let I = `int_0^(2π) (1)/(1 + e^(sin x)`dx ...(i)
Applying property,
`int_0^af(x)dx = int_0^af(a-x)dx,` we get
I = `int_0^(2pi) dx/(1+e^(sin(2pi-x)))`
= `int_0^(2pi)dx/(1+e^(-sinx))`
= `int_0^(2pi)dx/(1+1/e^(sinx))`
= `int_0^(2pi)(e^(sinx)dx)/(e^(sinx)+1)` ...(ii)
On adding equations (i) and (ii), we get
2I = `int_0^(2pi)dx/(1+e^(sinx))+int_0^(2pi)(e^(sinx)dx)/(1+e^(sinx))`
= `int_0^(2pi)((1+e^(sinx))/(1+e^(sinx)))dx`
= `int_0^(2pi)1.dx`
⇒ 2I = `[x]_0^(2pi)`
⇒ 2I = [2π]
⇒ I = π
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